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A coin is tossed thrice. i) Find the probability that there are two heads given that at least one is a tail. ii) Find the probability that there are two heads given that the first is a tail.

Mathematics

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2 parts
i

37\frac{3}{7}

ii

14\frac{1}{4}

Step-by-step solution

6 steps

First, let's list all possible outcomes when a coin is tossed thrice. The sample space SS is:

S={HHH,HHT,HTH,THH,HTT,THT,TTH,TTT}S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}

The total number of outcomes is ∣S∣=8|S| = 8.

Let AA be the event that there are exactly two heads.

A={HHT,HTH,THH}A = \{HHT, HTH, THH\}

So, ∣A∣=3|A| = 3.

Part i· 3 steps

Find the probability that there are two heads given that at least one is a tail.

  1. Define the event "at least one is a tail".

    Let BB be the event that at least one is a tail. This means any outcome except HHHHHH.

    B={HHT,HTH,THH,HTT,THT,TTH,TTT}B = \{HHT, HTH, THH, HTT, THT, TTH, TTT\}

    So, ∣B∣=7|B| = 7.

  2. Find the intersection of events AA and BB.

    The outcomes that are in both AA (exactly two heads) and BB (at least one tail) are:

    A∩B={HHT,HTH,THH}A \cap B = \{HHT, HTH, THH\}

    So, ∣A∩B∣=3|A \cap B| = 3.

  3. Calculate the conditional probability P(A∣B)P(A|B).

    The formula for conditional probability is P(A∣B)=P(A∩B)P(B)P(A|B) = \frac{P(A \cap B)}{P(B)}.

    P(A∩B)=∣A∩B∣∣S∣=38P(A \cap B) = \frac{|A \cap B|}{|S|} = \frac{3}{8} P(B)=∣B∣∣S∣=78P(B) = \frac{|B|}{|S|} = \frac{7}{8} P(A∣B)=3878=37P(A|B) = \frac{\frac{3}{8}}{\frac{7}{8}} = \frac{3}{7}

    The probability is 37\boxed{\frac{3}{7}}.

Part ii

14\frac{1}{4}

Find the probability that there are two heads given that the first is a tail.

  1. Define the event "the first is a tail".

    Let CC be the event that the first toss is a tail.

    C={THH,THT,TTH,TTT}C = \{THH, THT, TTH, TTT\}

    So, ∣C∣=4|C| = 4.

  2. Find the intersection of events AA and CC.

    The outcomes that are in both AA (exactly two heads) and CC (first is a tail) are:

    A∩C={THH}A \cap C = \{THH\}

    So, ∣A∩C∣=1|A \cap C| = 1.

  3. Calculate the conditional probability P(A∣C)P(A|C).

    P(A∩C)=∣A∩C∣∣S∣=18P(A \cap C) = \frac{|A \cap C|}{|S|} = \frac{1}{8} P(C)=∣C∣∣S∣=48=12P(C) = \frac{|C|}{|S|} = \frac{4}{8} = \frac{1}{2} P(A∣C)=1848=14P(A|C) = \frac{\frac{1}{8}}{\frac{4}{8}} = \frac{1}{4}

    The probability is 14\boxed{\frac{1}{4}}.

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